Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & {\text{As}}\,\,P \propto \frac{1}{{\;{V^{1.5}}}},\,{\text{So}}\,\,P{V^{1.5}} = {\text{constant}} \cr & \therefore \gamma = 1.5\,\,\left( {\because {\text{Process}}\,{\text{is}}\,{\text{adiabatic}}} \right) \cr & {\text{As}}\,{\text{we}}\,{\text{know,}}\,\,\frac{{{C_p}}}{{{C_v}}} = \gamma \cr & \therefore \frac{{{C_p}}}{{{C_v}}} = 1.5 \cr} $$

For path $$AB,$$  applying first law of thermodynamics
$$\eqalign{ & dQ = dU + dW\,\,{\text{or}}\,\,dQ = {U_B} - {U_A} + dW \cr & {\text{or}}\,\,600 = {U_B} - {U_A} + 0\,\,\left( {{\text{for}}\,{\text{isochoric}}\,{\text{process}}\,dV = 0\,\,{\text{so,}}\,\,dW = 0} \right) \cr & {U_B} - {U_A} = 600\,......\left( {\text{i}} \right) \cr} $$
For path $$BC,$$
$$\eqalign{ & dQ = dU + dW \cr & \therefore 200 = {U_C} - {U_B} + {p_B}\left( {{V_C} - {V_B}} \right)\,\,\left( {{\text{as}}\,BC\,{\text{is}}\,{\text{isobaric}}\,{\text{process}}} \right) \cr & \therefore 200 = {U_C} - {U_B} + 8 \times {10^4}\left( {5 \times {{10}^{ - 3}} - 2 \times {{10}^{ - 3}}} \right) \cr & {\text{or}}\,\,200 = {U_C} - {U_B} + 240 \cr & \therefore {U_C} - {U_B} = - 40\;\,J\,......\left( {{\text{ii}}} \right) \cr} $$
For $$AC,$$  change in internal energy can be calculated by adding Eqs. (i) and (ii),
$$\eqalign{ & = {U_C} - {U_A} = {U_C} - {U_B} + {U_B} - {U_A} \cr & = - 40 + 600 = 560\,J \cr} $$

Heat required for raising the temperature of the whole body by $$1°C$$  is called the thermal capacity of the body.

In an adiabatic process
$$\eqalign{ & T{V^{\gamma - 1}} = {\text{Constant}} \cr & {\text{or, }}{T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1} \cr} $$
For monoatomic gas $$\gamma = \frac{5}{3}$$
$$\eqalign{ & \left( {300} \right){V^{\frac{2}{3}}} = {T_2}{\left( {2\,V} \right)^{\frac{2}{3}}} \cr & \Rightarrow \,\,{T_2} = \frac{{300}}{{{{\left( 2 \right)}^{\frac{2}{3}}}}} \cr} $$
$${T_2} = 189K$$   (final temperature)
Change in internal energy $$\Delta U = n\frac{f}{2}R\,\Delta T$$
$$\eqalign{ & = 2\left( {\frac{3}{2}} \right)\left( {\frac{{25}}{3}} \right)\left( { - 111} \right) \cr & = - 2.7\,kJ \cr} $$

$$\eqalign{ & {\text{Here}}\,T{V^{\gamma - 1}} = {\text{constant}} \cr & {\text{As}}\,\gamma = \frac{5}{3},\,{\text{hence}}\,T{V^{\frac{2}{3}}} = {\text{constant}} \cr & {\text{Now}}\,{T_1}L_1^{\frac{2}{3}} = {T_2}L_2^{\frac{2}{3}}\,\,\left( {\because V \propto L} \right); \cr & {\text{Hence,}}\,\,\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^{2/3}} \cr} $$

For a cyclic process, $$\Delta U = 0\,\,{\text{or}}\,\,E = 0$$
NOTE
For any cyclic process, change in internal energy is always equal to zero and work done is equal to area under the cycle.

Number of moles of first gas $$ = \frac{{{n_1}}}{{{N_A}}}$$
Number of moles of second gas $$ = \frac{{{n_2}}}{{{N_A}}}$$
Number of moles of third gas $$ = \frac{{{n_3}}}{{{N_A}}}$$
If there is no loss of energy then
$$\eqalign{ & {P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV \cr & \frac{{{n_1}}}{{{N_A}}}R{T_1} + \frac{{{n_2}}}{{{N_A}}}R{T_2} + \frac{{{n_3}}}{{{N_A}}}R{T_3} \cr & = \frac{{{n_1} + {n_2} + {n_3}}}{{{N_A}}}R{T_{mix}} \cr & {T_{mix}} = \frac{{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}{{{n_1} + {n_2} + {n_3}}} \cr} $$

According to first law of thermodynamics $$\Delta Q = U + W$$
or $$\Delta U = Q - W$$
$$\Delta U =$$  change in internal energy
$$Q =$$  heat given to system
$$W =$$  work done
$$\eqalign{ & \therefore \Delta U = 2 \times 4.2 \times 1000 - 500 \cr & = 8400 - 500 = 7900\,J \cr} $$

All reversible engines working for the same temperature of source and sink have same efficiencies. If the temperatures are different, the efficiency is different.

$$\eqalign{ & dU = dQ - dW = \left( {8 \times {{10}^5} - 6.5 \times {{10}^5}} \right) \cr & = 1.5 \times {10^5}J \cr & dW = dQ - dU = {10^5} - 1.5 \times {10^5} \cr & = - 0.5 \times {10^5}J \cr} $$
$$-ve$$  sign indicates that work done on the gas is $$0.5 \times {10^5}J.$$